30. To evaluate the field using Gauss’ law, we employ a cylindrical surface of area 2πr Lwhere Lis very large (large enough that contributions from the ends of the cylinder become irrelevant to the calculation). The volume within this surface is V= πr2L, or expressed more appropriate to our needs: dV= 2πr L dr. The charge enclosed is, with 652.5 10 C/mA−=×, 24enc02.2rqArrLdrALrπ=π=³By Gauss’ law, we find
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